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Math Help - Second Derivative Test w/ no critical numbers

  1. #1
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    Second Derivative Test w/ no critical numbers

    Hello,

    Here is a problem that I'm really struggling with: f(x)=2-2x-x^3. F prime of this function is -2-3x^2 and when setting to zero to find critical numbers, you end up taking the sq. root of -2/3. So there are no critical numbers.

    I'm supposed to determine concavity and find inflection points. Per the second derivative test rules, F prime c must equal 0 and f double prime must exist near c. Since F prime C doesn't equal zero, as there are no critical numbers, is the graph also not concave up/down and are there also no inflection points? Am I getting this right?

    I'm so confused, Thank you.
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  2. #2
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    Quote Originally Posted by iJenn View Post
    Hello,

    Here is a problem that I'm really struggling with: f(x)=2-2x-x^3. F prime of this function is -2-3x^2 and when setting to zero to find critical numbers, you end up taking the sq. root of -2/3. So there are no critical numbers.

    I'm supposed to determine concavity and find inflection points. Per the second derivative test rules, F prime c must equal 0 and f double prime must exist near c. Since F prime C doesn't equal zero, as there are no critical numbers, is the graph also not concave up/down and are there also no inflection points? Am I getting this right?

    I'm so confused, Thank you.
    Application of the definitions:

    Concave up: f'' > 0.

    Concave down: f'' < 0.

    The function is a cubic and therefore has a point of inflection - the x-coordinate is therefore found by solving f'' = 0.
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    So the fact that f prime has no real critical numbers is irrelevant to the second derivative test for determining concavity?

    F double prime's critical point is (0,2) and is also the point of inflection. Is this correct?
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  4. #4
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    Quote Originally Posted by iJenn View Post
    So the fact that f prime has no real critical numbers is irrelevant to the second derivative test for determining concavity?

    F double prime's critical point is (0,2) and is also the point of inflection. Is this correct?
    Yes. Yes. A simple sketch graph of f(x) = 2 - 2x - x^3 confirms all this: plot y &#x3d;2-2x-x&#x5e;3 - Wolfram|Alpha
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